$f(x) \leq g(x), \forall x\in (a,\infty)$, prove that lim$_{x\rightarrow\infty} f\le\lim_{x\rightarrow\infty} g$ $f(x) \leq g(x), \forall x\in (a,\infty)$, prove that $\lim_{x\rightarrow\infty} f\leq$ lim$_{x\rightarrow\infty} g$ (Given both these limits exist)
My progress: Let $\lim_{x\rightarrow\infty} f=L$, and $\lim_{x\rightarrow\infty} g=M$.
$\therefore$ Given $\varepsilon>0$, $\exists k_1>a$ such that $|f(x)-L|<\varepsilon, \forall x>k_1$ and $\exists k_2>a$ such that $|g(x)-M|<\varepsilon, \forall x>k_2$
Let $k=\max\{ k_1,k_2\} \Rightarrow \forall x>k, |f(x)-L|<\varepsilon,|g(x)-M|<\varepsilon$
I need to show that $L\leq M$. I don't know how to go from here. Please help!
 A: Try with $\epsilon = |M - L|  / 2$.
A: Suppose $f(x)\geq0$ $\forall$ $x$.  Suppose $\lim\limits_{x\to\infty}f(x)=L<0$.  Let $\epsilon=|L|/2$.  Then $\exists$ $N$ such that $x>N$ $\Rightarrow$ $|f(x)-L|<|L|/2$.  This is equivalent to
$$L/2 < f(x)-L < -L/2$$
Thus $x>N$ $\Rightarrow$ $f(x) < L/2 < 0$.
Contradiction.
Now apply this to $g(x)-f(x)$.
A: It is much simpler to argue via contradiction. I use your notation of $L, M$ and on the contrary assume that $L > M$. This means that $L - M > 0$ and hence we can choose a positive number $\epsilon = (L - M)/2$.
Since $\lim\limits_{x \to \infty}f(x) = L$ it follows that there is a number $k_{1} > 0$ such that $$L - \epsilon < f(x) < L + \epsilon$$ whenever $x > k_{1}$. Thus we have $$f(x) > \frac{L + M}{2}\tag{1}$$ whenever $x > k_{1}$.
Again $\lim\limits_{x \to \infty}g(x) = M$ implies the existence of a number $k_{2} > 0$ such that $$M - \epsilon < g(x) < M + \epsilon$$ whenever $x > k_{2}$. And then we have $$g(x) < \frac{L + M}{2}\tag{2}$$ whenever $x > k_{2}$.
If we take $k$ as the maximum of $k_{1}, k_{2}, a$ and keep $x > k$ then we have both $x > k_{1}$ and $x > k_{2}$ and hence both inequalities $(1)$ and $(2)$ are satisfied simultaneously when $x > k$. We then have $$g(x) < \frac{L + M}{2} < f(x)$$ for all $x > k \geq a$. But this contradicts the given assumption that $f(x) \leq g(x)$ for all $x \geq a$.

How does one arrive at such a proof? The idea is simple. If $\lim\limits_{x \to \infty} f(x) = L$ and $\lim\limits_{x \to \infty} g(x) = M$ then $f(x)$ is near $L$ and $g(x)$ is near $M$ for all sufficiently large values of $x$. If $L > M$ then there will come a situation when $f(x)$ is so near to $L$ as to exceed $M$ and $g(x)$ is so near to $M$ as to be exceeded by $L$. The situation on number line will turn out to be like
-------|--------M--------|-------$\dfrac{L + M}{2}$-------||-------L--------||-------
Here single "|" represents the zone of values of $g(x)$ near $M$ and double "||" represents the zone of values of $f(x)$ near $L$. And it is clear that in this case $g(x) < f(x)$. In order that the zone of $g(x)$ remains separate from the zone of $f(x)$, it is necessary to find the mid point $(L + M)/2$ of $L, M$ on number line and ensure that the zone of $g(x)$ is left of this midpoint and zone of $f(x)$ is to the right of this midpoint.
In order to achieve this it is necessary to keep the distance between $g(x)$ and $M$ to be less than the distance of $M$ from this midpoint value $(L + M)/2$ and the required distance is $(L - M)/2$. Same goes for distance between $f(x)$ and $L$. This is how we arrive at $\epsilon = (L - M)/2$ and formalize (and sometimes intimidate a beginner learning calculus) a simple argument about inequalities.
